HDU 3038 How Many Answers Are Wrong (带权并查集)

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https://cn.vjudge.net/contest/66964#problem/D

带权并查集讲解博客：https://blog.csdn.net/sunmaoxiang/article/details/80959300

除了带权并查集，这道题还涉及到了区间问题，为了保证覆盖到每一个点，我们设置给定区间为左开右闭

这里权值的思想和向量运算很像（就是），利用了向量的有向性保证了运算的正确性

对于给定区间（u, v），我们默认向量方向为u -> v，赋予了方向以后，这道题就可做了

#include<iostream>
#include<cstring>
#include<cstdio>
#include<algorithm>
#include<vector>
#include<set>
#include<map>
#include<queue>
#include<cmath>
#define ll long long
#define mod 1000000007
#define inf 0x3f3f3f3f
using namespace std;
const int maxn = 200005;
const int maxm = 400005;
int pre[maxn], val[maxn];
//if(pre[i] == i) return i;
//else return find(pre[i]);
int find(int x)
{
    if(pre[x] != x)
    {
        int tmp = pre[x];
        pre[x] = find(pre[x]);
        val[x] += val[tmp];
    }
    return pre[x];
}
int main()
{
    int n, m;
    while(scanf("%d%d", &n, &m) != EOF)
    {
        for(int i = 0; i <= n; i ++)
        {
            pre[i] = i;
            val[i] = 0;
        }
        int u, v, w;
        int ans = 0;
        for(int i = 1; i <= m; i ++)
        {
            scanf("%d%d%d", &u, &v, &w);
            u --;
            int fu = find(u), fv = find(v);
            if(fu == fv && ((val[u] - val[v]) != w))
                ans ++;
            else if(fu != fv)
            {
                pre[fu] = fv;
                val[fu] = w + val[v] - val[u];
            }
        }
        printf("%d\n", ans);
    }
    return 0;
}